smurfy:

Not bad. Here goes...
Each side has eight pawns, and each pawn can move in 2! ways. So white's first pawn move can be done in 2! * 8 ways (= 16). Each side has two knights and each knight can move in 2! ways, giving 2! * 2 (=4). So white's first two moves can be done in 16+4=20 ways. Since each side has exactly the same number of ways of moving, then, using product principle, the first two moves can be made in 20*20=400 ways. By the way, are u on chess.com? ...feel like trashing someone right now lol

@Smurfy, chess.com?...i've not registerd yet...i'll do dat @ dusk so we can have a nice time...

Alpha Maximus: ....I can't believe no-one has attempted this question, guyz in the house, you done fall my hand, sorry, calculator

...@Maximus, d rational mathematician...till dusk...u'll c d solution...

rhydex 247: @ jackpot i tried to modify my error yesterday nite but d network hook me. Thanks 4 d correctn.
. . .

you're welcome, always

Laplacian:
...@Maximus, d rational mathematician...till dusk...u'll c d solution...

my friend, the rational Alpha Maximus no even decode the sarcasm sef

...@Maximux...suppose there are n people lockd in d conference room...d first man shakes n-1 ppl & he is then shut outside...second man shakes n-2...and d we have d follwn progresn afta each hanshak...n-1, n-2, n-3,....,2, 1 so dat d total handshakes is d sum of d first n-1 positv integers
=(n-1)n/2...equat and find n....
pls dont ask me 4 d value 'cos i've no calc on me...

Laplacian:
@Smurfy, chess.com?...i've not registerd yet...i'll do dat @ dusk so we can have a nice time...

Nice time? Did you just say nice time? Well, maybe getting trashed by ME is nice time to YOU...lol (I get mouth sha...)

2nioshine: @Jackpot u seems like my one tym gal who actually gave me a tough time and made me study like c.obi(hope u r nt d unizik gal)...alpha i luv ur math spirit no doubt u r no baby mathematician so lets b a little r***tional here(seun no dy pay or give us grades abeg),...double dx i was kinda di***pointed,but now u apologised we r kul...u remainent my most respectd boss...# Still in my observation mood#....nice wuk @ all 4 ur tym nd contributn(wish i could list al ur names)..sti following...wait o! Where is Richez

Boss seems richiez is busy these days, he is now one of the moderators in the education sector, I have been following this thread from the inception. Boss it been a long time we've seen your handiwork on mathematics oooooo

smurfy:

Nice time? Did you just say nice time? Well, maybe getting trashed by ME is nice time to YOU...lol (I get mouth sha...)

am set...we 'll c abt dat

Laplacian: ...@Maximux...suppose there are n people lockd in d conference room...d first man shakes n-1 ppl & he is then shut outside...second man shakes n-2...and d we have d follwn progresn afta each hanshak...n-1, n-2, n-3,....,2, 1 so dat d total handshakes is d sum of d first n-1 positv integers
=(n-1)n/2...equat and find n....
pls dont ask me 4 d value 'cos i've no calc on me...

..... Alright, that's actually a step in the right direction, but it still doesn't answer my question, and there has to be a calculator on whatever device you used to type. It doesn't have to be scientific, the question requires the use of only the BODMAS mathematical operations

Laplacian:
am set...we 'll c abt dat

Welcome on board, mate. My username is smurfy_chess. Search for it and invite for a game. Get ready to get smurfed!

@smurfy...i got dat...
@Maximus..ur solution comes in 5min's time

...@Maximus....for 55,130,250 handshakes, there are 10,501 people in d conferenc room, while for 31,988,001 handshakes, there are 7,999 people...10,501-7,999=2,502people

Laplacian: ...@Maximus....for 55,130,250 handshakes, there are 10,501 people in d conferenc room, while for 31,988,001 handshakes, there are 7,999 people...10,501-7,999=2,502people

Thumbs up man!! You got it right!!! Word problems are very fascinating branches of math and could be misleading and annoying at times!!! I'll be back with more. Seriously, you guyz should keep this stuff lively.

...sum d series;
cos@+cos2@+cos3@+...

...show that d intersection of any subgroups of a group G is a subgroup of G...

...G is a group of order 4, show dat G is abelian...

...questn1; find a number which wen multiplied by 7 and d result added to 1 gives a perfect square
....questn2; List ALL d subgroups of a group of order 7
...questn3; find All integers satisfying; 3x+4y=5
....question4; black king on a8, white king on a6, white pawn on c7...white to move and mate in 1

Alpha Maximus: I'm calculatorless, scientifically ..

but electronically u 're no...lol!!

Laplacian: ...questn1; find a number which wen multiplied by 7 and d result added to 1 gives a perfect square
....questn2; List ALL d subgroups of a group of order 7
...questn3; find All integers satisfying; 3x+4y=5
....question4; black king on a8, white king on a6, white pawn on c7...white to move and mate in 1

pawn advances to c8, gets promoted to Queen (or Rook) ...checkmate!

@Laplacian
Answer to que1 is 0.
So, what's ur username on chess.com?

smurfy: @Laplacian
Answer to que1 is 0.
So, what's ur username on chess.com?

answers mean nothin 2 me without steps...username is adokwu-ondoma

smurfy:

pawn advances to c8, gets promoted to Queen (or Rook) ...checkmate!

simple enough i guess

@ smurfy, wat devic do u use on chess.com?...cmp or mobil phon?

Laplacian: ...show that d intersection of any subgroups of a group G is a subgroup of G...

solution.
Suppose we have 2 subgroups of G defined as H1 and H2. Nw the questn implies H1 intersectn H2 must also be a subgroup of G.
For any element a in H1 there exists a^-1 and H2 is closed. The same holds for H2. So the intersectn will only contain an element c in H1 intersectn H2. If c and c^-1 are in H1 and H2 they must contain e the identity of G thus H1 intersectn H2 cannot be empty. Hence the intersectn of any subgroups of a group G is a subgroup of G.

Laplacian: @ smurfy, wat devic do u use on chess.com?...cmp or mobil phon?

Mobile phone.

Laplacian:
simple enough i guess

Yep. Such are meant for babies.